ICAR AICE-JRF/SRF (PhD admission and fellowship) · India · National Testing Agency for ICAR
Original practice questions written to the published syllabus · not past papersThe economics of turning inputs into farm output: production functions, the three basic relationships, optimum input use, efficiency, scale and risk.
In the exams: agricultural master's entrance examinations, factor-product relationships (heavy); agricultural economics PhD entrance and fellowship examinations, production economics (heavy); agricultural scientist eligibility tests (heavy)
ICAR AICE-JRF/SRF Agricultural Economics, Unit 5 (Agricultural Production Economics)
Algebra of production function forms (Cobb-Douglas, CES, transcendental, translog), returns to scale and Euler's theorem, cost minimisation with Lagrangian, isoclines and ridge lines, Farrell efficiency, SFA and DEA, Hicksian technical change and TFP, risk and utility, household models and duality lemmas.
How our questions for ICAR AICE-JRF/SRF spread across the units of Agricultural Production Economics. Difficulty: 2 easy, 27 medium, 28 hard.
| Duality among production, cost and profit functions | 9 | |
| Technical and allocative efficiency and production frontiers | 7 | |
| Forms of production function and their estimation | 6 | |
| Types of risk and enterprise combination under risk | 6 | |
| Theory of the family farm | 6 | |
| Cost minimisation | 4 | |
| Technological change | 4 | |
| Optimum input use: MVP equals MFC | 3 | |
| Economies of scale and size | 3 | |
| Law of variable proportions and the three stages of production | 2 | |
| Factor-factor relationship | 2 | |
| Isoquants and isoclines | 2 | |
| Scope of production economics compared with farm management | 1 | |
| Factor-product relationship | 1 | |
| Product-product relationship | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
Under constant returns to scale, stage I for the variable input corresponds to which stage for the fixed input?
A. Stage III, where the fixed input's marginal product is negative
With linear homogeneity the two stages are symmetric: where the variable input is too scarce relative to the fixed one (stage I), the fixed input is in excess and its marginal product is negative (its stage III). This is a second reason stage I is irrational.
Further reading: Doll and Orazem, Production Economics: Theory with Applications
With Y = 10X^0.5, output price 2 and input price 1, the profit-maximising input level is
A. 100 units
MPP = 5X^-0.5, so MVP = 2 x 5/sqrt(X) = 10/sqrt(X). Set equal to MFC = 1: sqrt(X) = 10, X = 100. Stopping at sqrt(X) gives 10; forgetting the 0.5 in MPP gives 400. Ignoring the output price (setting MPP = 1) gives 25.
Further reading: Debertin, Agricultural Production Economics
For Y = A X1^0.4 X2^0.2, with P1 = 50 and P2 = 100, the cost-minimising input ratio X2/X1 is
B. 0.25
Tangency: (b1/b2)(X2/X1) = P1/P2, so X2/X1 = (b2/b1)(P1/P2) = (0.2/0.4)(50/100) = 0.5 x 0.5 = 0.25. Inverting the price ratio gives 1, inverting both ratios gives 4, and ignoring prices gives 0.5. The answer does not depend on output, so the expansion path is a ray.
Further reading: Chambers, Applied Production Analysis: A Dual Approach
A farm's estimated Cobb-Douglas elasticities for all inputs sum to 1.15. What happens if one tries to find a profit maximum along the expansion path at constant prices?
C. No finite profit maximum exists, as marginal cost keeps falling with output
With increasing returns to scale the cost function is concave in output and marginal cost declines, so MC = P gives a profit minimum, not maximum. Cost minimisation is still possible for any given output; Cobb-Douglas has no output maximum.
Further reading: Beattie, Taylor and Watts, The Economics of Production
A production function is homogeneous of degree 0.8. If all inputs are doubled, output
A. rises by a factor of 1.741
Homogeneous of degree k: f(tX) = t^k f(X) = 2^0.8 = 1.741. Multiplying 2 x 0.8 = 1.6 is the common slip.
Further reading: Chambers, Applied Production Analysis: A Dual Approach
In the stochastic frontier model ln Y = f(X) + v - u, the component u represents
B. technical inefficiency, a one-sided non-negative term
v is symmetric noise; u >= 0 is inefficiency that pushes output below the frontier. TE = exp(-u). The model was proposed by Aigner, Lovell and Schmidt (1977) and Meeusen and van den Broeck (1977).
Further reading: Coelli, Rao, O'Donnell and Battese, An Introduction to Efficiency and Productivity Analysis
Efficiency scores from a frontier analysis of 200 farms in one district are best interpreted as
C. relative to the best performers in that sample, not absolute measures of efficiency
Frontiers are built from the observed farms, so a score of 1.0 means best among those observed. Scores from different samples or methods are not directly comparable.
Further reading: Coelli, Rao, O'Donnell and Battese, An Introduction to Efficiency and Productivity Analysis
Which property does the profit function pi(p, w) have?
D. Convex in prices and homogeneous of degree one in all prices together
Profit rises with output price, falls with input prices, is convex in prices (the firm adjusts to price changes, so profit rises at least linearly) and is linearly homogeneous. Input and output quantities, not profit, are homogeneous of degree zero in prices.
Further reading: Chambers, Applied Production Analysis: A Dual Approach
A cost function is C = 2y(w1 w2)^0.5. At y = 10, w1 = 4 and w2 = 9, the cost-minimising demand for input 1 is
B. 15
By Shephard's lemma, x1 = dC/dw1 = 2y x 0.5 x w1^-0.5 w2^0.5 = y(w2/w1)^0.5 = 10 x (9/4)^0.5 = 10 x 1.5 = 15. 120 is total cost; inverting the price ratio gives 6.667; forgetting the 0.5 from differentiation gives 30.
Further reading: Chambers, Applied Production Analysis: A Dual Approach
A single-output, single-input farm has profit function pi(p, w) = p^2/(4w). At p = 12 and w = 3, the profit-maximising output is
C. 2
Hotelling: y = d pi/dp = 2p/(4w) = p/(2w) = 12/6 = 2. 12 is profit itself; 4 is the input demand x = -d pi/dw = p^2/(4w^2). 1 drops the 2 from differentiating p^2 (p/(4w)).
Further reading: Chambers, Applied Production Analysis: A Dual Approach
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